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Note that the electric field is defined for a positive test charge q size 12{q} {} , so that the field lines point away from a positive charge and toward a negative charge. (See [link] .) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the electric field for a point charge is E = k | Q | / r 2 size 12{E= { ital "kQ"} slash {r rSup { size 8{2} } } } {} and area is proportional to r 2 size 12{r rSup { size 8{2} } } {} . This pictorial representation, in which field lines represent the direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields: electrostatic, gravitational, magnetic, and others.

In part a, electric field lines emanating from a positive charge is shown by the vector arrows in all direction of two dimensional space and the density of these field lines is less. In part b, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is less. In part c, electric field lines entering the negative charge is shown by the vector arrows coming from all direction of two dimensional space and the density of these field lines is large.
The electric field surrounding three different point charges. (a) A positive charge. (b) A negative charge of equal magnitude. (c) A larger negative charge.

In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created by each charge. The following example shows how to add electric field vectors.

Adding electric fields

Find the magnitude and direction of the total electric field due to the two point charges, q 1 size 12{q rSub { size 8{1} } } {} and q 2 size 12{q rSub { size 8{2} } } {} , at the origin of the coordinate system as shown in [link] .

Two charges are placed on a coordinate axes. Q two is at the position x equals 4 and y equals 0 centimeters. Q one is at the position x equals 0 and y equals two centimeters. Charge on q one is plus five point zero nano coulomb and charge on q two is plus ten nano coulomb. The electric field, E one having a magnitude of one point one three multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along positive y axis starting from the origin. The electric field, E two having magnitude zero point five six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow along negative x axis starting from the origin. The resultant field makes an angle of sixty three point four degree above the negative y axis having magnitude one point two six multiplied by ten raise to the power five Newton per coulomb is represented by a vector arrow pointing away from the origin in the second quadrant.
The electric fields E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} at the origin O add to E tot size 12{E rSub { size 8{"tot"} } } {} .

Strategy

Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, q size 12{q} {} , at point O, which allows us to determine the direction of the fields E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} . Once those fields are found, the total field can be determined using vector addition    .

Solution

The electric field strength at the origin due to q 1 size 12{q rSub { size 8{1} } } {} is labeled E 1 size 12{E rSub { size 8{1} } } {} and is calculated:

E 1 = k q 1 r 1 2 = 8 . 99 × 10 9 N m 2 /C 2 5 . 00 × 10 9 C 2 . 00 × 10 2 m 2 E 1 = 1 . 124 × 10 5 N/C . alignl { stack { size 12{E rSub { size 8{1} } =k { {q rSub { size 8{1} } } over {r rSub { size 8{1} } rSup { size 8{2} } } } = left (9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } right ) { { left (5 "." "00" times "10" rSup { size 8{ - 9} } C right )} over { left (2 "." "00" times "10" rSup { size 8{ - 2} } m right ) rSup { size 8{2} } } } } {} #E rSub { size 8{1} } =1 "." "125" times "10" rSup { size 8{5} } "N/C" {} } } {}

Similarly, E 2 size 12{E rSub { size 8{2} } } {} is

E 2 = k q 2 r 2 2 = 8 . 99 × 10 9 N m 2 /C 2 10 . 0 × 10 9 C 4 . 00 × 10 2 m 2 E 2 = 0 . 5619 × 10 5 N/C . alignl { stack { size 12{E rSub { size 8{2} } =k { {q rSub { size 8{2} } } over {r rSub { size 8{2} } rSup { size 8{2} } } } = left (9 "." "00" times "10" rSup { size 8{9} } N cdot m rSup { size 8{2} } "/C" rSup { size 8{2} } right ) { { left ("10" "." 0 times "10" rSup { size 8{ - 9} } C right )} over { left (4 "." "00" times "10" rSup { size 8{ - 2} } m right ) rSup { size 8{2} } } } } {} #E rSub { size 8{2} } =0 "." "5625" times "10" rSup { size 8{5} } "N/C" {} } } {}

Four digits have been retained in this solution to illustrate that E 1 size 12{E rSub { size 8{1} } } {} is exactly twice the magnitude of E 2 size 12{E rSub { size 8{2} } } {} . Now arrows are drawn to represent the magnitudes and directions of E 1 size 12{E rSub { size 8{1} } } {} and E 2 size 12{E rSub { size 8{2} } } {} . (See [link] .) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for E 1 size 12{E rSub { size 8{1} } } {} is exactly twice the length of that for E 2 size 12{E rSub { size 8{2} } } {} . The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field E tot size 12{E rSub { size 8{"tot"} } } {} is

E tot = ( E 1 2 + E 2 2 ) 1/2 = { ( 1.124 × 10 5 N/C ) 2 + ( 0.5619 × 10 5 N/C ) 2 } 1/2 = 1.26 × 10 5 N/C. alignl { stack { size 12{E rSub { size 8{ ital "tot"} } `= \( E rSub { size 8{1} } rSup { size 8{2} } `+`E rSub { size 8{2} } rSup { size 8{2} } \) rSup { size 8{ {1} wideslash {2} } } } {} #~``=` lbrace \( 1 "." "125" times "10" rSup { size 8{5} } `"N/C" \) rSup { size 8{2} } `+` \( 0 "." "5625" times "10" rSup { size 8{5} } `"N/C" \) rSup { size 8{2} } rbrace rSup { size 8{ {1} wideslash {2} } } {} # `~`=``1 "." "26" times "10" rSup { size 8{5} } `"N/C" {}} } {}

The direction is

θ = tan 1 E 1 E 2 = tan 1 1 . 124 × 10 5 N/C 0 . 5619 × 10 5 N/C = 63 . , alignl { stack { size 12{θ="tan" rSup { size 8{ - 1} } left ( { {E rSub { size 8{1} } } over {E rSub { size 8{2} } } } right )} {} #="tan" rSup { size 8{ - 1} } left lbrace { {1 "." "125" times "10" rSup { size 8{5} } " N/C"} over {0 "." "5625" times "10" rSup { size 8{5} } " N/C"} } right rbrace {} # ="63" "." 4° {}} } {}

or 63.4º above the x -axis.

Discussion

In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.

Questions & Answers

Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
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ok we can say body is electrically neutral ...conductor this quality is given to most metalls who have free electron in orbital d ...but human doesn't have ...so we re made from insulator or dielectric material ... furthermore, the menirals in our body like k, Fe , cu , zn
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Practice Key Terms 4

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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